Question

Find the value of d for an A.P., if t₅ = 11 and t₆ = 13...



please tell anyone plzzxxx​

Answer 1

Answer:

let d = common difference

6th term - 5th term

Step-by-step explanation:

13 - 11

2

Answer 2

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given

$t5 = 11 \\ t6 = 13$

$t5 = 11 \\ \implies \:t + 4d = 11 \: \: \: \: \: \: \: \: \: - eq.1 \\ and \\ t6 = 13 \\ \implies \: t + 5d = 13 \: \: \: \: \: \: \: \: \: - eq.2$

from equation 1

$t = 11 - 4d \: \: \: \: \: \: \: \: - eq.3$

now substitute the value of t in equation 2

$\implies \: t + 5d = 13 \\ \implies \: 11 - 4d + 5d = 13 \\ \implies \: d = 13 - 11 \\ \implies \: d = 2$

so common difference is 2

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if d = 2

and t5 = 11

$\implies \: t5 = 11 \\ \implies \: t + 4 \times 2 = 11 \\ \implies \: t = 11 - 8 \\ \implies \: t = 3$

so first term of AP is 3